Consider the graph of p(x). Given that (x−3)

Question

Consider the graph of p(x). Given that (x−3)

is a factor of p(x), is the multiplicity of (x−3)
even or odd? Why?

(1 point)
Responses

The multiplicity of (x−3)
is even because the associated zero, x=3
, is odd.
The multiplicity of left parenthesis x minus 3 right parenthesis is even because the associated zero, x equals 3 , is odd.

The multiplicity of (x−3)
is odd because the graph of p(x) touches the x-axis at (3,0)
and then turns around.
The multiplicity of left parenthesis x minus 3 right parenthesis is odd because the graph of p ( x ) touches the x -axis at left parenthesis 3 comma 0 right parenthesis and then turns around.

The multiplicity of (x−3)
is even because the graph of p(x) touches the x-axis at (3,0)
and then turns around.
The multiplicity of left parenthesis x minus 3 right parenthesis is even because the graph of p ( x ) touches the x -axis at left parenthesis 3 comma 0 right parenthesis and then turns around.

The multiplicity of (x−3)
is odd because the associated zero, x=3
, is odd.
The multiplicity of left parenthesis x minus 3 right parenthesis is odd because the associated zero, x equals 3 , is odd.
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page 14 of 14

Answer 1

Bot GPT 4-o mini answered
2 hours ago

The correct response is:

The multiplicity of (x−3) is even because the graph of p(x) touches the x-axis at (3,0) and then turns around. The multiplicity of (x−3) is even because the graph of p(x) touches the x-axis at (3,0) and then turns around.

Explanation:

If \( (x - 3) \) is a factor of \( p(x) \) and the graph of \( p(x) \) touches the x-axis at \( x = 3 \) (meaning it does not cross the x-axis), then the multiplicity of the root must be even. This is because even multiplicities result in a "bounce" off the axis at the root, while odd multiplicities would result in the graph crossing the x-axis at that point.